# Identifying series and parallel connections

• gracy
In summary: Yes, you are correct. My apologies for the confusion. I was not thinking clearly when I wrote that. You are right that the blue components are not in series because there are more than two components connected to that wire segment.
gracy said:
If there is 4 ohm resistance in an intended or normal path of a circuit and there is an optional path where 3 ohm resistance is present.Then current will take route where there is 3 ohm resistance.Will it be called short circuit or greater difference in resistance is required?
If the 3 Ohm resistance was not an intentional part of the original circuit then it is technically a short circuit.

If someone were to tell you to place a short circuit across the 4 Ohm path without any other details, then you would place zero Ohms (a wire) across it, because that is the conventional implication.

I can't provide a definition for "low" here. It depends on too many arbitrary or unspecified factors.

gracy
Ok.Thanks for all the answers @gneill .God bless you,give you all happiness in your life.Take care.
You made my day!I was having great difficulty in identifying parallel and series connections.Hope ,this thread helps many students who are facing same problem.

Here wires are arranged like a square

In this only second and third capacitors (from right)are in series.If these square like arrangement of wire is not there ALL the capacitors are in series.Here I have not used wires of different color that I should have used.

Here I could not find any capacitors in parallel as there are no capacitors sharing two nodes.But according to my book three capacitors are in parallel.The book however does not mention which three.

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You haven't identified the nodes properly. Any wire segments that touch must be the same node (color). In your diagram I can see orange touching blue, red touching green, green touching pink. Make another attempt.

gracy
When drawing a node, example the green note, keep on following the lines until block by resistor, capacitor or any other components. Trace all branches till blocked.

gneill said:
You haven't identified the nodes properly. Any wire segments that touch must be the same node (color). In your diagram I can see orange touching blue, red touching green, green touching pink. Make another attempt.
If I will replace blue with red,it will be wrong as red will be crossing the capacitor then.

Why didn't you replace blue with orange? If you started coloring on the left end with orange, you should stay with orange until you run out of wires that are connected. When you switch to red for the wire between the two capacitors on the left, stay with red until all wire segments that are connected to it are the same color, and so on.

gracy
I think this would be correct

And if we look at the above pic,it is clear that first three capacitors from left (from end p)are parallel to each other as the share 2 common nodes orange and red.
Right?
But we can see there was orange wire then a capacitor came in between then there was red and again orange,Will it be okay?

gracy said:
And if we look at the above pic,it is clear that first three capacitors from left (from end p)are parallel to each other as the share 2 common nodes orange and red.
Right?
Right.
But we can see there was orange wire then a capacitor came in between then there was red and again orange,Will it be okay?
Yes it's fine. So long as you identified the nodes properly (which you did this time) then whatever sequence appears must be correct.

If you had some other circuit where you discovered that both leads of a component ended up with the same color, then you would have to worry. The circuit would either be "wrong" or it's a trick question, since that component would be "shorted out" by the wiring and would not be doing anything useful in the circuit.

How can I simplify this circuit I mean how can I remove the square like arrangement of wires how to replace the square like arrangement of wires with some simple wiring?

gracy said:
How can I simplify this circuit I mean how can I remove the square like arrangement of wires how to replace the square like arrangement of wires with some simple wiring?
1. Identify the parallel components.
2. For each set of parallel components that you identify erase all but one and replace that one with the parallel value.
3. Clean up superfluous wires.

gracy
But one square is still remaining!

If I will replace the black wire with yellow then this remaining one wire will be not there.And all the three capacitors would be in parallel.Resultant capacitance would be 5C(All six capacitors had capacitance C)
gneill said:
3. Clean up superfluous wires.
I think by this you meant use least number of wires that's why I replaced black with yellow wire.

I didn't see your intermediate steps but I think something went wrong, probably an inadvertent change in the circuit when you were reducing the wiring. That 3c capacitor should not have ended up where you show it, and if it did it would be in parallel with the one beside it (so one or the other should not be there at this point).

Here's how I see the reduction taking place. I've replaced the orange color with green for better contrast with the red:

gracy
Thanks infinitely for your diagram.Neat and tidy!
I did not understand What exactly we have to erase?

In my book answer is given to be ##\frac{3C}{4}##
And explanation is given as follows
First three capacitors are in parallel and next two are short circuited.
I don't know if there is no trace of resistance how did they know it is short circuited?

gracy said:
Thanks infinitely for your diagram.Neat and tidy!
I did not understand What exactly we have to erase?
Erase all but one capacitor in each parallel grouping.

gracy
gracy said:
In my book answer is given to be ##\frac{3C}{4}##
And explanation is given as follows
First three capacitors are in parallel and next two are short circuited.
I don't know if there is no trace of resistance how did they know it is short circuited?
A wire is a equivalent to a zero valued resistance. The parallel pair on the right was shorted by the wire going over/around it. Using the method I've outlined this combined pair becomes an isolated capacitor automatically, so you don't need to do any other analysis to find the shorted (and thus redundant) components.

gracy
Note that "this capacitor "becomes isolated !so "so this parallel pair "was not doing anything in the circuit.
one of your lines from the immaculate diagram .

"this capacitor"
you meant capacitor having capacitance C (first capacitor from right hand side)?
"so this parallel pair "
Which pair?you meant
the second and third capacitors from right side.?
was not doing anything in the circuit.You meant there was no utility of them?
If yes,what does an isolated capacitor has to do with the pair's utility?

gracy said:
Note that "this capacitor "becomes isolated !so "so this parallel pair "was not doing anything in the circuit.
one of your lines from the immaculate diagram .

"this capacitor"
you meant capacitor having capacitance C (first capacitor from right hand side)?
No. Why would you think that? The note in the diagram points to the isolated capacitor which was formed from the parallel pair at that location. The far right capacitor was never touched by the reductions as it was not part of a parallel group.
"so this parallel pair "
Which pair?you meant
the second and third capacitors from right side.?
Of course. It's a pair of capacitors in the original circuit, and we found them to be parallel.
was not doing anything in the circuit.You meant there was no utility of them?
If yes,what does an isolated capacitor has to do with the pair's utility?
The parallel pair was reduced to one capacitor. The reduction process left it with only one lead connected to the circuit.

Without both leads connected a component cannot carry current. Charge cannot move onto or off of a capacitor through one wire alone. The capacitor thus has no influence, no utility, as far as the circuit is concerned. This means that the original pair was also of no utility (a correct circuit simplification cannot change the circuit's externally-viewed behavior in any way).

gracy
gneill said:
A wire is a equivalent to a zero valued resistance
Wires will always be there in a circuit,then this means all the circuits short circuited.Of course not,what am I missing?

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I now understand how tha answer came out to be ##\frac{3C}{4}##
But I am still facing difficulty in comprehending the following
gneill said:
A wire is a equivalent to a zero valued resistance. The parallel pair on the right was shorted by the wire going over/around it.

gracy said:
Wires will always be there in a circuit,then won't this all the circuits are short circuited?Of course not,what am I missing?
Go back to the discussion of what a "short circuit" means. It means a bypass (either intentional or accidental) of a normal circuit path. Wires (paths) that are supposed to be there to interconnect components are just doing what they were meant to do.

In the circuit under discussion two of the capacitors are bypassed by a wire. Whether this was intentional (i.e. a "trick" for you to discover) or an accident (the author of the problem made a mistake) cannot be known for sure, but because it is not a proper situation to have for a practical circuit (why would one design a circuit with parts that do nothing?) we call that path a short circuit: the subcircuit comprising the two capacitors is shorted. Current will not flow in that subcircuit due to the bypass.

gracy

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